PENG GAO
Abstract. We use a theorem of Cartlidge and the technique of Redheffer’s ”recurrent inequalities” to give some results on inequalities related to Hardy’s inequality.
1. Introduction
Suppose throughout that p6= 0,1p +1q = 1. Let lp be the Banach space of all complex sequences a= (an)n≥1 with norm
||a||:= (
∞
X
n=1
|an|p)1/p <∞.
The celebrated Hardy’s inequality([10], Theorem 326) asserts that forp >1,
(1.1)
∞
X
n=1 (1
n
n X
k=1
ak)p ≤(
p p−1)
p
∞
X
k=1
|ak|p.
Among the many papers appeared providing new proofs, generalizations and sharpenings of (1.1), we refer the reader to the work of G.Bennett [2]-[6] for his study of factorable matrices.
Hardy’s inequality can be regarded as a special case of the following inequality:
∞
X
j=1
∞
X
k=1
cj,kak p
≤U ∞
X
k=1
|ak|p,
in whichC = (cj,k) and the parameterp are assumed fixed(p >1), and the estimate is to hold for all real sequences a. The lp operator norm of C is then defined as the p-th root of the smallest value of the constantU:
||C||p,p =U 1
p.
Hardy’s inequality thus asserts that theCes´aro matrix operatorC, given bycj,k = 1/j, k≤j and 0 otherwise, is bounded onlp and has norm ≤p/(p−1). (The norm is in factp/(p−1).)
We say a matrix A is a summability matrix if its entries satisfy: aj,k ≥0,aj,k = 0 for k > j and Pj
k=1aj,k = 1. We say a summability matrix A is a weighted mean matrix if its entries satisfy:
(1.2) aj,k =λk/Λj, 1≤k≤j; Λj = j X
i=1
λi.
We refer to the n-tuple (an1, an2,· · · , ann) as the n-th row of a summability matrix Aand then have the following result of Bennett([6], Theorem 1.14) for thelp operator normof A.
Theorem 1.1. Let p > 1 be fixed and suppose A is a summability matrix. If the rows of A are decreasing, then ||A||p,p ≥p/(p−1). If the rows of A are increasing, then ||A||p,p ≤p/(p−1).
The above theorem, when applied to weighted mean matrixes, gives the following inequality([6], Corollary 4.10).
Date: Auguest 7, 2003.
1991Mathematics Subject Classification. Primary 26D15,26D20.
Key words and phrases. Hardy’s inequality, Carleman’s inequality, recurrent inequality.
Theorem 1.2. If 0< λ1 ≤λ2≤ · · · and 0< p <1, then (1.3)
∞
X
n=1 (
Pn i=1λiapi Pn
i=1λi
)1/p≤( 1 1−p)
1/p
∞
X
n=1
an,
whenever a is a sequence of non-negative terms.
Even though the constant in the above theorem is best possible, some improvement may be possible with specific choices of theλi’s. For examples, the following two inequalities were claimed to hold by Bennett([5], page 40-41; see also [6], page 407):
∞
X
n=1 ( 1
nα n X
i=1
(iα−(i−1)α)ai)p ≤ (
αp αp−1)
p
∞
X
n=1
|an|p, (1.4)
∞
X
n=1 (Pn 1
i=1iα−1 n X
i=1
iα−1ai)p ≤ (
αp αp−1)
p
∞
X
n=1
|an|p, (1.5)
whenever α >0, p >1, αp >1.
We haven’t seen the proofs of Bennett but find the following unpublished result of J. Cartlidge[7] is very helpful to treat the above two inequalities. We don’t have access to his thesis either, so here we quote the one in [2](p. 416):
Theorem 1.3. Let 1< p <∞ be fixed. Let A be a weighted mean matrix given by (1.2). If
(1.6) L= sup
n
(Λn+1
λn+1
−Λn λn
)< p ,
then||A||p,p≤p/(p−L).
We will apply the above theorem to prove (1.4)-(1.5) for α≥2, p >1, αp >1 in section 3. Supposean ≥0, by a change of variables an→ a1/pn and let p→ ∞, (1.1) gives the well-known Carleman’s inequality:
∞
X
n=1 (
n Y
k=1
ak) 1
n ≤e
∞
X
n=1
an.
We refer the reader to the survey article [13] and the references therein for an account of Carle-man’s inequality. Among the various generalizations of CarleCarle-man’s inequality, we mention a result of E. Love, who proved forα >0, λi =iα−(i−1)α,
(1.7)
∞
X
n=1 (
n Y
i=1
aiiα−(i−1)α)1/nα ≤eα1
∞
X
n=1
an,
and the constant eα1 is best possible. We note here after a change of variables an → a1/pn , (1.7)
corresponds to the limiting case p→ ∞ of (1.4).
R.Redheffer gave a remarkable proof of Hardy’s inequality in [14] by developing the method of ”recurrent inequalities”. His method also works for Carleman’s inequality. Another proof of Carleman’s inequality was given by him in [15] and his result has been generalized by H.Alzer[1] and most recently by J. Peˇcari´c and K. Stolarsky[13], who proved forbn>0,N ≥1, Gn= (Qni=1ai)1/n,
N X
n=1
Λn(bn−1)Gn+ ΛNGN ≤ N X
n=1
λnGnbΛnn/λn.
In this paper, we will use Redheffer’s method to give a weighted version of his treatment of Hardy’s and Carleman’s inequalities. As we shall see, our result for 1< p <∞ is less satisfactory than that of Cartlidge’s while for the limiting case the result is almost the same as his.
2. Lemmas
Lemma 2.1. Let Λk = Pki=1λi, λi > 0 and Sn = Pni=1λiai. Let 0 6= p < 1 be fixed and let (µn)n≥1,(ηn)n≥1 be two sequences of real numbers such that µi ≤ηi for 0< p <1 and µi ≥ηi for
p <0, then for n≥2,
(2.1)
n−1 X
i=2
[µi−(µqi+1−η q i+1)
1/q]S1/p
i +µnSn1/p≤(µ q 2−η
q 2)
1/qλ1/p 1 a
1/p 1 +
n X
i=2
ηiλ1/pi a 1/p i .
Proof. This is essentially due to R.Redheffer[14]. We note fork≥2,
(2.2) µkSk1/p−ηkλ1/pk a1/pk =Sk1/p−1(µk(1 +t) 1/p−η
it1/p)≤(µqk−ηkq)1/qS1/pk−1,
with t =λkak/Sk−1(compare this with the one on page 688 of [14]). The lemma then follows by
adding (2.2) for 2≤k≤ntogether.
Lemma 2.2. Let Λk =Pki=1λi, λi >0 and Gk= (Qki=1aλii)1/Λk, then forµi >0, n≥2,
(2.3) G1+
n−1 X
i=2 (Λiµi
λi
− Λi λi+1
)Gi+ Λnµn
λn
Gn≤(1 + Λ1
λ2 )a1+
n X
i=2
µ
Λi
λi
i ai.
Proof. This is essentially due to R.Redheffer[14]. We note fork≥2, µ >0, η >0,
µGk−ηak=Gk−1(µt−ηt Λk
λk)≤Gk−1(Λk−1
λk )η
−λk
Λk−1(µλk Λk
) Λk Λk−1,
where t
Λk
λk = ak/Gk−1(compare this with the one on page 686 of [14]). By setting µkΛk/λk =
µ, ηk =η=µ Λk/λk
k , we get
(2.4) Λkµk
λk
Gk−akµ Λk
λk
k ≤ Λk−1
λk
Gk.
The lemma then follows by adding (2.4) for 2≤k≤nand G1 =a1 together. Lemma 2.3. Let f(x)∈C3[a, b] andf000(x)≥0 for x∈[a, b]. Then
(2.5) f(b)−f(a)≥f0(a+b
2 )(b−a).
Proof. By Taylor’s expansion,
f(b) = f(a+b 2 ) +f
0(a+b
2 )(b−
a+b
2 ) +f
00(η
1)(a−b)2/4,
f(a) = f(a+b 2 ) +f
0(a+b
2 )(a−
a+b
2 ) +f
00(η
2)(a−b)2/4,
wherea < η2 <(a+b)/2< η1< b. The lemma then follows by noticingf000(x)≥0 forx∈[a, b]. Lemma 2.4. If s≥1, then
(2.6)
n X
i=1
is≥ s s+ 1
ns(n+ 1)s (n+ 1)s−ns.
3. Applications of Cartlidge’s Theorem
We say a weighted mean matrix A given by (1.2) is generated by a logarithmico-exponential function if for all sufficiently large n,λn:=l(n), where l(x) is a positive logarithmico-exponential function and a logarithmico-exponential function on [x0,∞] is defined by Hardy[9] as a real valued function defined by a finite combination of ordinary algebraic symbols(viz, +,−,×,÷,√n) and the
functional symbolslog(·) ande(·), operating on real variablex and on real constants. We note first the following theorem of F. Cass and W. Kratz[8]:
Theorem 3.1. Let 1< p <∞ be fixed. Let A be a weighted mean matrix given by (1.2). Suppose
limn→∞Λn/nλn=L < p, then p/(p−L)≤ ||A||p,p.
It is easy to see limn→∞nα−1/(nα−(n−1)α) = 1/αand the simplest Euler-Maclaurin formulae
gives:
n X
i=1
f(i) = Z n
1
f(x)dx+f(1) + Z n
1
(x−[x])f0(x)dx,
for f having continuous derivative f0, where [x] denote the largest integer not exceeding the real number x. It then follows
n X
i=1
iα−1 =nα/α+o(nα).
Thus thanks to Theorem 3.1, we know if (1.4)-(1.5) hold for some α >0, p >1, αp >1 then the constants (αp/(αp−1))p are best possible.
Now we apply Cartlidge’s Theorem to get
Corollary 3.1. Inequality (1.4) holds for p > 1, α ≥ 2, αp > 1 and the constant there is best possible.
Proof. Apply Theorem 1.3 withλi=iα−(i−1)α. We definef(x) =xα/(xα−(x−1)α), x≥1 so that Λi+1/λi+1−Λi/λi =f(i+ 1)−f(i) =f0(ξ),1≤i < ξ < i+ 1, with
0< f0(ξ) = αξ
α−1(ξ−1)α−1 (ξα−(ξ−1)α)2 ≤
1
α,
where the last inequality follows from Lemma 2.3 and the arithmetic-geometric inequality, since for
α≥2,
ξα−(ξ−1)α ≥α(ξ+ (ξ−1)
2 )
α−1 ≥α(ξ(ξ−1))(α−1)/2.
This completes the proof.
We note the corollary implies (1.7) forα≥2. Now if we apply Theorem 1.3 to (1.5), we need to show
n+1 X
i=1
iα−1/(n+ 1)α−1−
n X
i=1
iα−1/nα−1 = 1 + ( 1
(n+ 1)α−1 − 1
nα−1) n X
i=1
iα−1 ≤1/α.
The second inequality above follows from Lemma 2.4 and we get
4. Generalizations of Redheffer’s Results
Theorem 4.1. Assume the same conditions in Lemma 2.1 and let 0 < p < 1 be fixed. Suppose there exists a positive constant c such that c−1+ 1≤c−1/p and
(4.1) c≤1−p+ (1−p)(λ−i q−λ−i−q1)Λi−1λq/pi , i≥2.
Then for 0< p <1,
(4.2)
∞
X
i=1
(Si/Λi)1/p≤c−1/p
∞
X
i=1
a1/pi .
Proof. It suffices to prove the theorem for any integer n ≥1. We note first the condition (4.1) is equivalent to
(4.3) q−1(1−c−1+c−1Λi−1λq/pi (λ−i−q1−λ
−q
i ))≥1, i≥2. By setting ηi = λ
−1/p i , µ
q i = λ
−q/p
i + Λi−1/cλqi−1 in (2.1), we can rewrite the left-hand side of (2.1) as
(1−c−1/q)a1/p1 + n−1 X
i=2
[(λ−iq/p+ Λi−1/cλqi−1)1/q−(Λi/cλqi)1/q]S 1/p
i +µnSn1/p. By the mean value theorem,
(λ−i q/p+ Λi−1/cλqi−1)1/q−(Λi/cλqi)1/q ≥ q
−1(λ−q/p
i + Λi−1/cλqi−1−Λi/cλqi)(Λi/cλqi)
−1/p
= q−1(1−c−1+c−1Λi−1λq/pi (λ
−q i−1−λ
−q
i ))(Λi/c)−1/p
≥ (Λi/c)−1/p.
Here the last inequality follows from (4.3). Thus (2.1) becomes n
X
i=1
(Si/λi)1/p≤(c−1+ 1)a1+c−1/p n X
i=2
ai ≤c−1/p n X
i=1
ai.
This completes the proof.
We note here if 0 < λ1 ≤ λ2 ≤ · · ·, we can take c = 1−p in (4.1) and one checks easily for 0< p <1, (1−p)−1+ 1<(1−p)−1/p. Theorem 4.1 then implies Theorem 1.2.
We also note the constant given by the above theorem may be less satisfactory. For example the case α = 2, p= 2 in (1.4) corresponds to the case λi = 2i−1, p = 1/2, c= 3/4 in (4.2). However, direct calculation shows (4.1) is not satisfied in this case. Of course one may try to prove directly
(λ−i q/p+ Λi−1/cλqi−1)1/q−(Λi/cλqi)1/q ≥(Λi/c)−1/p. But one checks this fails fori= 2.
Similarly, the caseα = 2, p= 2 in (1.5) corresponds to the caseλi=i, p= 1/2, c= 3/4 in (4.2). One checks in this case (4) holds for i≥2. However,c−1+ 1 = 7/3>16/9 =c−2, so the coefficient of a1 is slightly larger.
Now we focus our attention to Carleman-type inequalities.
Theorem 4.2. Assume the same conditions in Lemma 2.2 and let f(x) be a real valued function defined forx≥2 such thatf(n) = Λn/λnfor n≥2 and0≤f(x+ 1)−f(x)≤1/αfor some α >0.
If (1 + Λ1λ2)≤e1/α for the same α then
(4.4)
∞
X
n=1 (
n Y
i=1
aλi
i )
1/Λn ≤(1 +Λ1
λ2 )a1+
n X
i=2
ai(1 +f(i+ 1)−f(i)
f(i) )
f(i) ≤e1/α
∞
X
n=1
Proof. It suffices to prove the theorem for any integern≥2. Setµi =f(i+ 1)/f(i) in Lemma 2.2 we get
n X
i=1
Gi ≤ n−1 X
i=1
Gi+f(n+ 1)GN ≤(1 + Λ1
λ2 )a1+
n X
i=2
ai(1 +f(i+ 1)−f(i)
f(i) )
f(i) ≤e1/α
∞
X
n=1
an,
by the conditions of the theorem and this completes the proof.
Apply Theorem 4.2 to λ1 = 1, λi =αi−1−αi−2, i≥ 2 for someα > 1, then f(x) = α/(α−1) and we get
Corollary 4.1. Forα >1,
(4.5)
∞
X
n=1 (a1
n Y
k=2
akαk−1−αk−2)1/αn−1 ≤(1 + 1
α−1)a1+
∞
X
n=2
an.
Apply Theorem 4.2 to λi =αi, i≥1 for some α >0, then f(i+ 1)−f(i) =α−i and we get Corollary 4.2. Forα >0,
(4.6)
∞
X
n=1 (
n Y
k=1
aαkk−1)(αn−1)/(α−1) ≤(1 + 1
α)a1+ ∞
X
n=2
e1/αnan≤
∞
X
n=1
e1/αnan.
We end the paper by noting that if we take λi = (i(i+ 1))−1 in Theorem 4.2, then f(x) = x2 and we get back a result of Redheffer(see [14]page 693):
Corollary 4.3. (4.7)
∞
X
n=1 (
n Y
k=1
a1/k(k+1))(n+1)/n≤ ∞
X
n=1
e2nan.
References
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Department of Mathematics, University of Michigan, Ann Arbor, MI 48109
